Integrand size = 29, antiderivative size = 65 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=-\frac {3 e^{\frac {5 (a-d)}{3}+\frac {2}{3} (d+b x)} (f-g)}{4 b}+\frac {3 e^{\frac {5 (a-d)}{3}+\frac {8}{3} (d+b x)} (f+g)}{16 b} \] Output:
-3/4*exp(5/3*a-d+2/3*b*x)*(f-g)/b+3/16*exp(5/3*a+d+8/3*b*x)*(f+g)/b
Time = 0.20 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=\frac {3 e^{\frac {5 a}{3}+\frac {2 b x}{3}} \left (\left (\left (-4+e^{2 b x}\right ) f+\left (4+e^{2 b x}\right ) g\right ) \cosh (d)+\left (\left (4+e^{2 b x}\right ) f+\left (-4+e^{2 b x}\right ) g\right ) \sinh (d)\right )}{16 b} \] Input:
Integrate[E^((5*(a + b*x))/3)*(g*Cosh[d + b*x] + f*Sinh[d + b*x]),x]
Output:
(3*E^((5*a)/3 + (2*b*x)/3)*(((-4 + E^(2*b*x))*f + (4 + E^(2*b*x))*g)*Cosh[ d] + ((4 + E^(2*b*x))*f + (-4 + E^(2*b*x))*g)*Sinh[d]))/(16*b)
Time = 0.23 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.72, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2720, 27, 802, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int e^{\frac {5}{3} (a+b x)} (f \sinh (b x+d)+g \cosh (b x+d)) \, dx\) |
\(\Big \downarrow \) 2720 |
\(\displaystyle \frac {3 \int -\frac {1}{2} e^{\frac {5 a}{3}+\frac {b x}{3}} \left (f-g-e^{2 b x} (f+g)\right )de^{\frac {b x}{3}}}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {3 e^{5 a/3} \int e^{\frac {b x}{3}} \left (f-g-e^{2 b x} (f+g)\right )de^{\frac {b x}{3}}}{2 b}\) |
\(\Big \downarrow \) 802 |
\(\displaystyle -\frac {3 e^{5 a/3} \int \left (e^{\frac {b x}{3}} (f-g)-e^{\frac {7 b x}{3}} (f+g)\right )de^{\frac {b x}{3}}}{2 b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {3 e^{5 a/3} \left (\frac {1}{2} e^{\frac {2 b x}{3}} (f-g)-\frac {1}{8} e^{\frac {8 b x}{3}} (f+g)\right )}{2 b}\) |
Input:
Int[E^((5*(a + b*x))/3)*(g*Cosh[d + b*x] + f*Sinh[d + b*x]),x]
Output:
(-3*E^((5*a)/3)*((E^((2*b*x)/3)*(f - g))/2 - (E^((8*b*x)/3)*(f + g))/8))/( 2*b)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[Exp andIntegrand[(c*x)^m*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]
Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct ionOfExponentialQ[u, x] && !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ [{a, m, n}, x] && IntegerQ[m*n]] && !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) *(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
Time = 1.69 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.08
method | result | size |
risch | \(-\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}} f}{4 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}-d +\frac {2 b x}{3}} g}{4 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}+d +\frac {8 b x}{3}} f}{16 b}+\frac {3 \,{\mathrm e}^{\frac {5 a}{3}+d +\frac {8 b x}{3}} g}{16 b}\) | \(70\) |
default | \(\frac {3 \left (-\frac {f}{2}+\frac {g}{2}\right ) \sinh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{2 b}+\frac {3 \left (-\frac {f}{2}+\frac {g}{2}\right ) \cosh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{2 b}+\frac {3 \left (\frac {g}{2}+\frac {f}{2}\right ) \sinh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{8 b}+\frac {3 \left (\frac {g}{2}+\frac {f}{2}\right ) \cosh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{8 b}\) | \(94\) |
orering | \(\frac {15 \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right )}{8 b}-\frac {9 \left (\frac {5 b \,{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g \cosh \left (b x +d \right )+f \sinh \left (b x +d \right )\right )}{3}+{\mathrm e}^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (g b \sinh \left (b x +d \right )+f b \cosh \left (b x +d \right )\right )\right )}{16 b^{2}}\) | \(97\) |
parts | \(f \left (-\frac {3 \sinh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{4 b}+\frac {3 \sinh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{16 b}-\frac {3 \cosh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{4 b}+\frac {3 \cosh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{16 b}\right )+g \left (\frac {3 \sinh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{4 b}+\frac {3 \sinh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{16 b}+\frac {3 \cosh \left (\frac {5 a}{3}-d +\frac {2 b x}{3}\right )}{4 b}+\frac {3 \cosh \left (\frac {5 a}{3}+d +\frac {8 b x}{3}\right )}{16 b}\right )\) | \(136\) |
Input:
int(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d)),x,method=_RETURNVERBO SE)
Output:
-3/4*exp(5/3*a-d+2/3*b*x)/b*f+3/4*exp(5/3*a-d+2/3*b*x)/b*g+3/16*exp(5/3*a+ d+8/3*b*x)/b*f+3/16*exp(5/3*a+d+8/3*b*x)/b*g
Leaf count of result is larger than twice the leaf count of optimal. 478 vs. \(2 (41) = 82\).
Time = 0.09 (sec) , antiderivative size = 478, normalized size of antiderivative = 7.35 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx =\text {Too large to display} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d)),x, algorithm="f ricas")
Output:
3/16*((f + g)*cosh(1/3*b*x + 1/3*d)^4*cosh(-5/3*a + 5/3*d) + ((f + g)*cosh (-5/3*a + 5/3*d) - (f + g)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^4 - 4*(f - g)*cosh(1/3*b*x + 1/3*d)^2*cosh(-5/3*a + 5/3*d) + 4*((f + g)*cosh( 1/3*b*x + 1/3*d)*cosh(-5/3*a + 5/3*d) - (f + g)*cosh(1/3*b*x + 1/3*d)*sinh (-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^3 + 2*(3*(f + g)*cosh(1/3*b*x + 1/ 3*d)^2*cosh(-5/3*a + 5/3*d) - 2*(f - g)*cosh(-5/3*a + 5/3*d) - (3*(f + g)* cosh(1/3*b*x + 1/3*d)^2 - 2*f + 2*g)*sinh(-5/3*a + 5/3*d))*sinh(1/3*b*x + 1/3*d)^2 + 4*((f + g)*cosh(1/3*b*x + 1/3*d)^3*cosh(-5/3*a + 5/3*d) + 2*(f - g)*cosh(1/3*b*x + 1/3*d)*cosh(-5/3*a + 5/3*d) - ((f + g)*cosh(1/3*b*x + 1/3*d)^3 + 2*(f - g)*cosh(1/3*b*x + 1/3*d))*sinh(-5/3*a + 5/3*d))*sinh(1/3 *b*x + 1/3*d) - ((f + g)*cosh(1/3*b*x + 1/3*d)^4 - 4*(f - g)*cosh(1/3*b*x + 1/3*d)^2)*sinh(-5/3*a + 5/3*d))/(b*cosh(1/3*b*x + 1/3*d)^4 - 4*b*cosh(1/ 3*b*x + 1/3*d)^3*sinh(1/3*b*x + 1/3*d) + 6*b*cosh(1/3*b*x + 1/3*d)^2*sinh( 1/3*b*x + 1/3*d)^2 - 4*b*cosh(1/3*b*x + 1/3*d)*sinh(1/3*b*x + 1/3*d)^3 + b *sinh(1/3*b*x + 1/3*d)^4)
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (46) = 92\).
Time = 0.23 (sec) , antiderivative size = 133, normalized size of antiderivative = 2.05 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=\begin {cases} \frac {15 f e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh {\left (b x + d \right )}}{16 b} - \frac {9 f e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \cosh {\left (b x + d \right )}}{16 b} - \frac {9 g e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \sinh {\left (b x + d \right )}}{16 b} + \frac {15 g e^{\frac {5 a}{3}} e^{\frac {5 b x}{3}} \cosh {\left (b x + d \right )}}{16 b} & \text {for}\: b \neq 0 \\x \left (f \sinh {\left (d \right )} + g \cosh {\left (d \right )}\right ) e^{\frac {5 a}{3}} & \text {otherwise} \end {cases} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d)),x)
Output:
Piecewise((15*f*exp(5*a/3)*exp(5*b*x/3)*sinh(b*x + d)/(16*b) - 9*f*exp(5*a /3)*exp(5*b*x/3)*cosh(b*x + d)/(16*b) - 9*g*exp(5*a/3)*exp(5*b*x/3)*sinh(b *x + d)/(16*b) + 15*g*exp(5*a/3)*exp(5*b*x/3)*cosh(b*x + d)/(16*b), Ne(b, 0)), (x*(f*sinh(d) + g*cosh(d))*exp(5*a/3), True))
Time = 0.04 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.09 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=\frac {3}{16} \, g {\left (\frac {e^{\left (\frac {8}{3} \, b x + \frac {5}{3} \, a + d\right )}}{b} + \frac {4 \, e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a - d\right )}}{b}\right )} + \frac {3}{16} \, f {\left (\frac {e^{\left (\frac {8}{3} \, b x + \frac {5}{3} \, a + d\right )}}{b} - \frac {4 \, e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a - d\right )}}{b}\right )} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d)),x, algorithm="m axima")
Output:
3/16*g*(e^(8/3*b*x + 5/3*a + d)/b + 4*e^(2/3*b*x + 5/3*a - d)/b) + 3/16*f* (e^(8/3*b*x + 5/3*a + d)/b - 4*e^(2/3*b*x + 5/3*a - d)/b)
Time = 0.12 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=\frac {3 \, {\left (f e^{\left (\frac {8}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )} + g e^{\left (\frac {8}{3} \, b x + \frac {5}{3} \, a + 2 \, d\right )} - 4 \, f e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a\right )} + 4 \, g e^{\left (\frac {2}{3} \, b x + \frac {5}{3} \, a\right )}\right )} e^{\left (-d\right )}}{16 \, b} \] Input:
integrate(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d)),x, algorithm="g iac")
Output:
3/16*(f*e^(8/3*b*x + 5/3*a + 2*d) + g*e^(8/3*b*x + 5/3*a + 2*d) - 4*f*e^(2 /3*b*x + 5/3*a) + 4*g*e^(2/3*b*x + 5/3*a))*e^(-d)/b
Time = 2.84 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.71 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=\frac {3\,{\mathrm {e}}^{\frac {5\,a}{3}-d+\frac {2\,b\,x}{3}}\,\left (4\,g-4\,f+f\,{\mathrm {e}}^{2\,d+2\,b\,x}+g\,{\mathrm {e}}^{2\,d+2\,b\,x}\right )}{16\,b} \] Input:
int(exp((5*a)/3 + (5*b*x)/3)*(g*cosh(d + b*x) + f*sinh(d + b*x)),x)
Output:
(3*exp((5*a)/3 - d + (2*b*x)/3)*(4*g - 4*f + f*exp(2*d + 2*b*x) + g*exp(2* d + 2*b*x)))/(16*b)
Time = 0.23 (sec) , antiderivative size = 52, normalized size of antiderivative = 0.80 \[ \int e^{\frac {5}{3} (a+b x)} (g \cosh (d+b x)+f \sinh (d+b x)) \, dx=\frac {3 e^{\frac {5 b x}{3}+\frac {5 a}{3}} \left (-3 \cosh \left (b x +d \right ) f +5 \cosh \left (b x +d \right ) g +5 \sinh \left (b x +d \right ) f -3 \sinh \left (b x +d \right ) g \right )}{16 b} \] Input:
int(exp(5/3*b*x+5/3*a)*(g*cosh(b*x+d)+f*sinh(b*x+d)),x)
Output:
(3*e**((5*a + 5*b*x)/3)*( - 3*cosh(b*x + d)*f + 5*cosh(b*x + d)*g + 5*sinh (b*x + d)*f - 3*sinh(b*x + d)*g))/(16*b)